Given a string,determine if a permutation of the string Could form a palindrome.
For example,"code"
-> False, "aab"
-> True, "carerac"
-> True.
Hint:
- Consider the palindromes of odd vs even length. What difference do you notice?
- Count the frequency of each character.
- If each character occurs even number of times,then it must be a palindrome. How about character which occurs odd number of times?
给定一个字符串,确定该字符串的排列是否可以形成回文。
例如,
“code”->false,“aab”->true,“carerac”->true。
提示:
- 考虑奇数和偶数的回文长度。你注意到了什么区别?
- 计算每个字符的频率。
- 如果每个字符出现偶数次,那么它必须是回文。奇数次出现的字符怎么样?
1 class Solution { 2 func canPermutepalindrome(_ s:String) -> Bool { 3 var t:Set<Character> = Set<Character>() 4 for a in s.characters 5 { 6 if !t.contains(a) 7 { 8 t.insert(a) 9 } 10 else 11 { 12 t.remove(a) 13 } 14 } 15 return t.isEmpty || t.count == 1 16 } 17 }
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